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Transcript
26. SOLVING A SYSTEM OF LINEAR EQUATIONS 1) Must have both equations in the form ax + by = c : "2x = !3y + 15 "2x + 3y = 15 %# # 5x = 2y ! 29 $ $5x ! 2y = !29 2) We want to eliminate one of the variables. To do this, we need to take the first set of coefficients, flip them, and change the sign of one number. Then, we multiply both equations by those numbers. " 2x + 3y = 15 # $ 5x ! 2y = !29 2 5 " 5(2x + 3y = 15) # $!2(5x ! 2y = !29) 5 2 flip them change sign of one number 5 –2 " 10x + 15y = 75 # $!10x + 4 y = 58 19y = 133 19 19 y=7 3) Plug in the value of the variable you have just solved for into either of the original equations and solve for the other variable. To check, plug solutions for both variables into the remaining equation. y = 7 ! 2x = "3(7) + 15 2x = "21+ 15 2x = "6 2 2 x = "3 Visit SHAHOMEWORK.com Check (!3,7) 5x = 2y ! 29 5(!3) = 2(7) ! 29 !15 = 14 ! 29 √ !15 = !15 Created by J. Shahom (June 2004) Examples 1. ! a + 3b = 19 !1a + 3b = 19 $" " # a = 2b + 4 #1a % 2b = 4 !1(1a ! 2b = 4) " 1a + 3b = 19 # $!1a + 2b = !4 5b = 15 5 5 b=3 a + 3b = 19 a + 3(3) = 19 a + 9 = 19 !9 ! 9 a = 10 1(1a + 3b = 19) (10,3) Check 2. a = 2b + 4 10 = 2(3) + 4 10 = 6 + 4 √ 10 = 10 "2x = 14 ! 3y "2x + 3y = 14 %# # $5x ! 2y = !3 $5x ! 2y = !3 " 10x + 15y = 70 # $!10x + 4 y = 6 19y = 76 19 19 y=4 5(2x + 3y = 14) !2(5x ! 2y = !3) 2x = 14 ! 3y 2x = 14 ! 3(4) 2x = 14 !12 2x = 2 2 2 x =1 (1,4) Check Visit SHAHOMEWORK.com 5x ! 2y = !3 5(1) ! 2(4) = !3 5 ! 8 = !3 √ !3 = !3 Created by J. Shahom (June 2004)